1 \section{Single-domain proteins under constant loading}
3 Let $x$ be the end to end distance of the protein, $t$ be the time since loading began, $F$ be tension applied to the protein, $P$ be the surviving population of folded proteins.
6 v &\equiv \deriv{t}{x} && \text{the pulling velocity} \\
7 k &\equiv \deriv{x}{F} && \text{the loading spring constant} \\
8 P_0 &\equiv P(t=0) && \text{the initial number of folded proteins} \\
9 D &\equiv P_0 - P && \text{the number of dead (unfolded) proteins} \\
10 \kappa &\equiv -\frac{1}{P} \deriv{t}{P} && \text{the unfolding rate}
12 \nomenclature{$\equiv$}{Defined as (\ie equivalent to)}
13 The proteins are under constant loading because
15 \deriv{t}{F} = \deriv{x}{F}\deriv{t}{x} = kv\;,
17 a constant, since both $k$ and $v$ are constant (\citet{evans97} in the text on the first page, \citet{dudko06} in the text just before Eqn.~4).
19 The instantaneous likelyhood of a protein unfolding is given by $\deriv{F}{D}$, and the unfolding histogram is merely this function discretized over a bin of width $W$(This is similar to \citet{dudko06} Eqn.~2, remembering that $\dot{F}=kv$, that their probability density is not a histogram ($W=1$), and that their pdf is normalized to $N=1$).
21 h(F) \equiv \deriv{\text{bin}}{F}
22 = \deriv{F}{D} \cdot \deriv{\text{bin}}{F}
25 = -W \deriv{t}{P} \deriv{F}{t}
26 = \frac{W}{vk} P\kappa \label{eq:unfold:hist}
28 Solving for theoretical histograms is merely a question of taking your chosen $\kappa$, solving for $P(f)$, and plugging into Eqn. \ref{eq:unfold:hist}.
29 We can also make a bit of progress solving for $P$ in terms of $\kappa$ as follows:
31 \kappa &\equiv -\frac{1}{P} \deriv{t}{P} \\
32 -\kappa \dd t \cdot \deriv{t}{F} &= \frac{\dd P}{P} \\
33 \frac{-1}{kv} \int \kappa \dd F &= \ln(P) + c \\
34 P &= C\exp{\p({\frac{-1}{kv}\integral{}{}{F}{\kappa}})} \;, \label{eq:P}
36 where $c \equiv \ln(C)$ is a constant of integration scaling $P$.
38 \subsection{Constant unfolding rate}
40 In the extremely weak tension regime, the proteins' unfolding rate is independent of tension, we have
42 P &= C\exp{\p({\frac{-1}{kv}\integral{}{}{F}{\kappa}})}
43 = C\exp{\p({\frac{-1}{kv}\kappa F})}
44 = C\exp{\p({\frac{-\kappa F}{kv}})} \\
45 P(0) &\equiv P_0 = C\exp(0) = C \\
46 h(F) &= \frac{W}{vk} P \kappa
47 = \frac{W\kappa P_0}{vk} \exp{\p({\frac{-\kappa F}{kv}})}
49 So, a constant unfolding-rate/hazard-function gives exponential decay.
50 Not the most earth shattering result, but it's a comforting first step, and it does show explicitly the dependence in terms of the various unfolding-specific parameters.
52 \subsection{Bell model}
54 Stepping up the intensity a bit, we come to Bell's model for unfolding
55 (\citet{hummer03} Eqn.~1 and the first paragraph of \citet{dudko06} and \citet{dudko07}).
57 \kappa = \kappa_0 \cdot \exp\p({\frac{F \dd x}{k_B T}})
58 = \kappa_0 \cdot \exp(a F) \;,
60 where we've defined $a \equiv \dd x/k_B T$ to bundle some constants together.
61 The unfolding histogram is then given by
63 P &= C\exp\p({\frac{-1}{kv}\integral{}{}{F}{\kappa}})
64 = C\exp\p[{\frac{-1}{kv} \frac{\kappa_0}{a} \exp(a F)}]
65 = C\exp\p[{\frac{-\kappa_0}{akv}\exp(a F)}] \\
66 P(0) &\equiv P_0 = C\exp\p({\frac{-\kappa_0}{akv}}) \\
67 C &= P_0 \exp\p({\frac{\kappa_0}{akv}}) \\
68 P &= P_0 \exp\p\{{\frac{\kappa_0}{akv}[1-\exp(a F)]}\} \\
69 h(F) &= \frac{W}{vk} P \kappa
70 = \frac{W}{vk} P_0 \exp\p\{{\frac{\kappa_0}{akv}[1-\exp(a F)]}\} \kappa_0 \exp(a F)
71 = \frac{W\kappa_0 P_0}{vk} \exp\p\{{a F + \frac{\kappa_0}{akv}[1-\exp(a F)]}\} \label{eq:unfold:bell_pdf}\;.
73 The $F$ dependent behavior reduces to
75 h(F) \propto \exp\p[{a F - b\exp(a F)}] \;,
77 where $b \equiv \kappa_0/akv \equiv \kappa_0 k_B T / k v \dd x$ is
78 another constant rephrasing.
80 This looks similar to the Gompertz / Gumbel / Fisher-Tippett
83 p(x) &\propto z\exp(-z) \\
84 z &\equiv \exp\p({-\frac{x-\mu}{\beta}}) \;,
88 p(x) \propto z\exp(-bz) \;.
90 Strangely, the Gumbel distribution is supposed to derive from an
91 exponentially increasing hazard function, which is where we started
92 for our derivation. I haven't been able to find a good explaination
93 of this discrepancy yet, but I have found a source that echos my
94 result (\citet{wu04} Eqn.~1). TODO: compare \citet{wu04} with
95 my successful derivation in \cref{sec:sawsim:results-scaffold}.
97 Oh wait, we can do this:
99 p(x) \propto z\exp(-bz) = \frac{1}{b} z'\exp(-z')\propto z'\exp(-z') \;,
101 with $z'\equiv bz$. I feel silly... From
102 \href{Wolfram}{http://mathworld.wolfram.com/GumbelDistribution.html},
103 the mean of the Gumbel probability density
105 P(x) = \frac{1}{\beta} \exp\p[{\frac{x-\alpha}{\beta}
106 -\exp\p({\frac{x-\alpha}{\beta}})
109 is given by $\mu=\alpha-\gamma\beta$, and the variance is
110 $\sigma^2=\frac{1}{6}\pi^2\beta^2$, where $\gamma=0.57721566\ldots$ is
111 the Euler-Mascheroni constant. Selecting $\beta=1/a=k_BT/\dd x$,
112 $\alpha=-\beta\ln(\kappa\beta/kv)$, and $F=x$ we have
115 &= \frac{1}{\beta} \exp\p[{\frac{F+\beta\ln(\kappa\beta/kv)}{\beta}
116 -\exp\p({\frac{F+\beta\ln(\kappa\beta/kv)}
119 &= \frac{1}{\beta} \exp(F/\beta)\exp[\ln(\kappa\beta/kv)]
120 \exp\p\{{-\exp(F/\beta)\exp[\ln(\kappa\beta/kv)]}\} \\
121 &= \frac{1}{\beta} \frac{\kappa\beta}{kv} \exp(F/\beta)
122 \exp\p[{-\kappa\beta/kv\exp(F/\beta)}] \\
123 &= \frac{\kappa}{kv} \exp(F/\beta)\exp[-\kappa\beta/kv\exp(F/\beta)] \\
124 &= \frac{\kappa}{kv} \exp(F/\beta - \kappa\beta/kv\exp(F/\beta)] \\
125 &= \frac{\kappa}{kv} \exp(aF - \kappa/akv\exp(aF)] \\
126 &= \frac{\kappa}{kv} \exp(aF - b\exp(aF)]
129 So our unfolding force histogram for a single Bell domain under
130 constant loading does indeed follow the Gumbel distribution.
132 \subsection{Saddle-point Kramers' model}
134 For the saddle-point approximation for Kramers' model for unfolding
135 (\citet{evans97} Eqn.~3, \citet{hanggi90} Eqn. 4.56c, \citet{vanKampen07} Eqn. XIII.2.2).
137 \kappa = \frac{D}{l_b l_{ts}} \cdot \exp\p({\frac{-E_b(F)}{k_B T}}) \;,
139 where $E_b(F)$ is the barrier height under an external force $F$,
140 $D$ is the diffusion constant of the protein conformation along the reaction coordinate,
141 $l_b$ is the characteristic length of the bound state $l_b \equiv 1/\rho_b$,
142 $\rho_b$ is the density of states in the bound state, and
143 $l_{ts}$ is the characteristic length of the transition state
148 \citet{evans97} solved this unfolding rate for both inverse power law potentials and cusp potentials.
150 \subsubsection{Inverse power law potentials}
153 E(x) = \frac{-A}{x^n}
155 (e.g. $n=6$ for a van der Waals interaction, see \citet{evans97} in
156 the text on page 1544, in the first paragraph of the section
157 \emph{Dissociation under force from an inverse power law attraction}).
158 Evans then goes into diffusion constants that depend on the
159 protein's end to end distance, and I haven't worked out the math
163 \subsubsection{Cusp potentials}
166 E(x) = \frac{1}{2}\kappa_a \p({\frac{x}{x_a}})^2
168 (see \citet{evans97} in the text on page 1545, in the first paragraph
169 of the section \emph{Dissociation under force from a deep harmonic well}).